In another thread it was claimed that the operator $O : \operatorname{dom}(O) \subset L^2(-1,1) \rightarrow L^2(-1,1)$ is self-adjoint.
$$Of(x)= \frac{f(x)}{{1-x^2}}$$ It is obvious that $$\langle O f,g \rangle = \int_{-1}^{1} Ofg = \int_{-1}^{1} f \overline{Og} = \langle f,Og\rangle$$
This just shows symmetry and since $\operatorname{dom}(O) = \{f \in L^2: Of \in L^2\}$ is dense (contains all testfunctions) we just know from this that $O \subset O^*$.
So how can I prove the converse.
I'm going to build on another problem of yours, Tobias: Why is this operator self-adoint .
In the above problem, it is shown that, if $A : \mathcal{D}(A)\subseteq H\rightarrow H$ is symmetric with $(A\pm iI)$ surjective, then $A$ is densely-defined and selfadjoint.
As you noted, your operator $O$ is symmetric on its domain. To see that $(O\pm iI)$ are surjective, suppose $f \in L^{2}(-1,1)$ and define $$ g_{\pm} = \frac{(1-x^{2})f}{x\pm i}. $$ Clearly $g_{\pm}$ are in $L^{2}(-1,1)$, but also $g_{\pm}\in\mathcal{D}(O)$ with $$ (O\pm iI)g_{\pm} = f. $$ Conclusion: $\mathcal{O}$ is densely-defined and selfadjoint.